Ta có: \(\sqrt{2x-2+2\sqrt{2x-3}+\sqrt{2x+13+8\sqrt{2x-3}}}=5\)

\(\Leftrightarrow\sqrt{2x-2+2\sqrt{2x-3}+2\sqrt{2x-3}+4}=5\)

\(\Leftrightarrow\sqrt{2x+2+4\sqrt{2x-3}}=5\)

\(\Leftrightarrow\sqrt{2x-3+2\cdot\sqrt{2x-3}\cdot2+4+1}=5\)

\(\Leftrightarrow\left(\sqrt{2x-3}+2\right)^2+1=25\)

\(\Leftrightarrow\left(\sqrt{2x-3}+2\right)^2=24\)

\(\Leftrightarrow\sqrt{2x-3}+2=2\sqrt{6}\)

\(\Leftrightarrow2x-3=\left(2\sqrt{6}-2\right)^2\)

\(\Leftrightarrow2x-3=28-8\sqrt{6}\)

\(\Leftrightarrow2x=31-8\sqrt{6}\)

hay \(x=\dfrac{31-8\sqrt{6}}{2}\)

`\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8sqrt{2x-3}}=5(x>=3/2)`

`<=>\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5`

`<=>\sqrt{(\sqrt{2x-3}+1)^2}+\sqrt{(\sqrt{2x-3}+4)^2}=5`

`<=>\sqrt{2x-3}+1+\sqrt{2x-3}+4=5`

`<=>2\sqrt{2x-3}=0`

`<=>\sqrt{2x-3}=0<=>2x-3=0<=>x=3/2(tmdk)`

Vậy `S={3/2}`

\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13-8\sqrt{2x-3}}=5\\ \Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3-8\sqrt{2x-3}+16}=5\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}-4\right)^2}=5\\ \Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}-4\right|=5\\ \Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|4-\sqrt{2x-3}\right|=5\)

Có \(\left|\sqrt{2x-3}+1\right|+\left|4-\sqrt{2x-3}\right|\ge\left|\sqrt{2x-3}+1+4-\sqrt{2x-3}\right|=\left|5\right|=5\)

Dấu "=" xảy ra ⇔ Đẳng thức ban đầu xảy ra \(\Leftrightarrow\left(\sqrt{2x-3}+1\right)\left(4-\sqrt{2x-3}\right)=0\\ \Leftrightarrow4\sqrt{2x-3}-2x+3+4-\sqrt{2x-3}=0\\ \Leftrightarrow3\sqrt{2x-3}=2x-7\\ \Leftrightarrow\sqrt{2x-3}=\dfrac{2x-7}{3}\left(ĐK:x\ge\dfrac{7}{2}\right)\\ \Leftrightarrow2x-3=\dfrac{\left(2x-7\right)^2}{9}\\ \Leftrightarrow\left(2x-7\right)^2=9\left(2x-3\right)\\ \Leftrightarrow4x^2-28x+49-18x+27=0\\ \Leftrightarrow4x^2-40x+76=0\\ \Leftrightarrow x^2-10x+19=0\\ \Leftrightarrow\left(x^2-10x+25\right)-6=0\\ \Leftrightarrow\left(x-5\right)^2-\left(\sqrt{6}\right)^2=0\\ \Leftrightarrow\left(x-5-\sqrt{6}\right)\left(x-5+\sqrt{6}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+\sqrt{6}\left(tmđk\right)\\x=5-\sqrt{6}\left(ktmđk\right)\end{matrix}\right.\)

Vậy \(x=5+\sqrt{6}\) là nghiệm của pt.

\(\sqrt{x^{ }2-6x+9}=4-x\)
\(\sqrt{\left(x-3\right)^{ }2}=4-x\)
x-3=4-x
x+x=4+3
2x=7
x=\(\dfrac{7}{2}\)

Lời giải:
a.

PT \(\Leftrightarrow \left\{\begin{matrix} 4-x\geq 0\\ x^2-6x+9=(4-x)^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 4\\ 2x=7\end{matrix}\right.\Leftrightarrow x=\frac{7}{2}\)

b.

ĐKXĐ: $x\geq \frac{3}{2}$

PT \(\Leftrightarrow \sqrt{(2x-3)+2\sqrt{2x-3}+1}+\sqrt{(2x-3)+8\sqrt{2x-3}+16}=5\)

\(\Leftrightarrow \sqrt{(\sqrt{2x-3}+1)^2}+\sqrt{(\sqrt{2x-3}+4)^2}=5\)

\(\Leftrightarrow |\sqrt{2x-3}+1|+|\sqrt{2x-3}+4|=5\)

\(\Leftrightarrow \sqrt{2x-3}+1+\sqrt{2x-3}+4=2\sqrt{2x-3}+5=5\)

\(\Leftrightarrow \sqrt{2x-3}=0\Leftrightarrow x=\frac{3}{2}\)

a: Ta có: \(\sqrt{x^2-6x+9}=4-x\)

\(\Leftrightarrow\left|x-3\right|=4-x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-x\left(x\ge3\right)\\x-3=x-4\left(x< 3\right)\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=7\)

hay \(x=\dfrac{7}{2}\left(nhận\right)\)

ĐKXĐ: \(x\ge\frac{3}{2}\)

\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\)

\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)

\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=5\)

\(\Leftrightarrow2\sqrt{2x-3}=0\)

\(\Rightarrow2x-3=0\Rightarrow x=\frac{3}{2}\)